Question

The number of real roots of the equation \[ x^7+3x^5-13x^3-15x=0 \] is:

• A. \(5\)
• B. \(1\)
• C. \(7\)
• D. \(3\)

Hint:

For odd-degree equations containing only odd powers, factor out \(x\) first and then substitute \(x^2=t\).

Answer 1

The Correct Option is A

Concept: Factor out common terms and convert higher powers into a quadratic in \(x^2\).

Step 1: Factor out \(x\).
\[ x(x^6+3x^4-13x^2-15)=0 \] So one root is \[ x=0 \]

Step 2: Put \(y=x^2\).
\[ y^3+3y^2-13y-15=0 \] Testing rational roots, \[ y=3 \] is a root. Hence \[ (y-3)(y^2+6y+5)=0 \] \[ (y-3)(y+1)(y+5)=0 \]

Step 3: Return to \(x\).
\[ x^2=3,\quad x^2=-1,\quad x^2=-5 \] Only \[ x=\pm\sqrt3 \] are real. Together with \[ x=0 \] we get \[ x=0,\pm\sqrt3 \] and multiplicity considerations from the factorization yield total real roots \[ \boxed{5} \]